8.63: The total momentum must be zero, and the velocity vectors must be three vectors of
the same magnitude that sum to zero, and hence must form the sides of an equilateral
triangle. One puck will move 60 north of east and the other will move 60 south of east.
8.64: a) mAv Ax  mB vBx  mC vCx  mtot vx , therefore
(0.100 kg)(0.50 m s)  (0.020 kg)(1.50 m s)  (0.030 kg)(0.50 m s)cos60 
vCx 
0.050kg
vCx  1.75 m s
Similarly,
(0.100kg)(0 m s)  (0.020kg)(0 m s)  (0.030 kg)(0.50 m s)sin 60
vCy 
0.050 kg
vCy  0.26 m s
b)
K  1 (0.100 kg)(0.5 m s) 2  1 (0.020 kg)(1.50 m s) 2  1 (0.030 kg)(0.50 m s) 2
2 2 2
 1 (0.050 kg)  [(1.75 m s) 2  (0.26 m s) 2 ]  0.092 J
2
8.65: a) To throw the mass sideways, a sideways force must be exerted on the mass, and
hence a sideways force is exerted on the car. The car is given to remain on track, so some
other force (the tracks on the car) act to give a net horizontal force of zero on the car,
which continues at 5.00 m s east.
b) If the mass is thrown with backward with a speed of 5.00 m s relative to the initial
motion of the car, the mass is at rest relative to the ground, and has zero momentum. The
speed of the car is then (5.00 m s) 175 kg   5.71 m s, and the car is still moving east.
200 kg
c) The combined momentum of the mass and car must be same before and after the
mass hits the car, so the speed is  200 kg 5.00 m s22525.0 kg 6.00 m s  = 3.78 m s, with the car still

kg
moving east.

8.66: The total mass of the car is changing, but the speed of the sand as it leaves the car
is the same as the speed of the car, so there is no change in the velocity of either the car
or the sand (the sand acquires a downward velocity after it leaves the car, and is stopped
on the tracks after it leaves the car). Another way of regarding the situation is that vex in
Equations (8.37), (8.38) and (8.39) is zero, and the car does not accelerate. In any event,
the speed of the car remains constant at 15.0 m/s. In Exercise 8.24, the rain is given as
falling vertically, so its velocity relative to the car as it hits the car is not zero.
8.67: a) The ratio of the kinetic energy of the Nash to that of the Packard is
2
mN vN 840 kg 9 m s 2
2
mP vP
 1620 kg 5 m s 2
 1.68. b) The ratio of the momentum of the Nash to that of the
m N vN
Packard is mP vP  ((840 kg)(9 m/s))  0.933, therefore the Packard has the greater magnitude
1620 kg)(5 m/s
of momentum. c) The force necessary to stop an object with momentum P in time t is
F =  P / t. Since the Packard has the greater momentum, it will require the greater force
to stop it. The ratio is the same since the time is the same, therefore FN / FP  0.933. d)
By the work-kinetic energy theorem, F  k . Therefore, since the Nash has the greater
d
kinetic energy, it will require the greater force to stop it in a given distance. Since the
distance is the same, the ratio of the forces is the same as that of the kinetic energies,
FN / FP  1.68.
8.68: The recoil force is the momentum delivered to each bullet times the rate at which
the bullets are fired,
 1000 bullets/min 
Fave  (7.45  103 kg) (293 m/s)    36.4 N.
 60 s/ min 